This result is proved in [1]} in the case when \(B\)  and the subalgebras of \(\mathfrak {b}\)  are specified. We omit the proof since the argument still works in general because it uses only the assumptions that \(\mathfrak {n}^{\prime }+\mathfrak {h} = \mathfrak {b}\)  and \(\mathfrak {h}\) , \(\mathfrak {n}^{\prime }\) , \(\mathfrak {b}/\mathfrak {n}^{\prime }\)  are nilpotent.
